【Leetcode】17、Letter Combinations of a Phone Number
阿新 • • 發佈:2018-07-31
question pos dia ppi stage ble you xpl class
題目
Given a string containing digits from 2-9
inclusive, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Example:
Input: "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
思路
關鍵就是兩個循環,在原有X的基礎上,比如新加數字3,則Xd Xe Xf
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> result; if(digits.empty())return vector<string>(); vector<string> v = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; result.push_back(""); // add a seed for the initial case for(int i = 0 ; i < digits.size(); ++i) { int num = digits[i]-‘0‘;if(num < 0 || num > 9) break; string candidate = v[num]; if(candidate.empty()) continue; vector<string> tmp; for(int j = 0 ; j < candidate.size() ; ++j) { for(int k = 0 ; k < result.size() ; ++k) { tmp.push_back(result[k] + candidate[j]); } } result = tmp; } return result; } };
Explanation with sample input "123"
Initial state:
- result = {""}
Stage 1 for number "1":
- result has {""}
- candiate is "abc"
- generate three strings "" + "a", ""+"b", ""+"c" and put into tmp,
tmp = {"a", "b","c"} - swap result and tmp (swap does not take memory copy)
- Now result has {"a", "b", "c"}
Stage 2 for number "2":
- result has {"a", "b", "c"}
- candidate is "def"
- generate nine strings and put into tmp,
"a" + "d", "a"+"e", "a"+"f",
"b" + "d", "b"+"e", "b"+"f",
"c" + "d", "c"+"e", "c"+"f" - so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
- swap result and tmp
- Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
Stage 3 for number "3":
- result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
- candidate is "ghi"
- generate 27 strings and put into tmp,
- add "g" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- add "h" for each of "ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"
- so, tmp has
{"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
"adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
"adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" } - swap result and tmp
- Now result has
{"adg", "aeg", "afg", "bdg", "beg", "bfg", "cdg", "ceg", "cfg"
"adh", "aeh", "afh", "bdh", "beh", "bfh", "cdh", "ceh", "cfh"
"adi", "aei", "afi", "bdi", "bei", "bfi", "cdi", "cei", "cfi" }
Finally, return result.
【Leetcode】17、Letter Combinations of a Phone Number